To solve the differential equation y" + xy = 0 using a power series method, we can express y as a power series:
y = ∑(Ck * x^k)
where Ck represents the coefficient of the k-th power of x.
Given the substitution y = ∑(Ck * x^k), we want to find an expression for Ck+2 in terms of Ck-1 for k = 1, 2, 3, ...
To find Ck+2 in terms of Ck-1, we can differentiate y with respect to x twice and substitute it back into the original differential equation y" + xy = 0.
1. Differentiate y twice:
y' = ∑(k * Ck * x^(k-1))
y'' = ∑(k * (k-1) * Ck * x^(k-2))
2. Substitute y' and y'' back into the differential equation:
∑(k * (k-1) * Ck * x^(k-2)) + x * ∑(Ck * x^k) = 0
3. Multiply out the terms and adjust the indices to match Ck+2 and Ck-1:
∑((k+2) * (k+1) * Ck+2 * x^k) + x * ∑(Ck * x^k) = 0
4. Equate the coefficients of like powers of x:
(k+2) * (k+1) * Ck+2 + Ck = 0
5. Rearrange the equation to express Ck+2 in terms of Ck-1:
Ck+2 = - Ck / ((k+2) * (k+1))
By following these steps, you can find an expression for Ck+2 in terms of Ck-1 for the given differential equation when solving it using a power series method.
